I have a three part question, which I could only received an answer for the first part here.

1. The Laplace-Beltrami operator is an operator which is the typical example of a self-adjoint operator in $L^{2}$. I am wondering if this is also true for other Hilbert spaces $H^{k}$.
2. The second part of the question and in which I am more interested is the following: How much the results are true for metrics with are not positive definitive. For example, the D'Alambertian(wave operator) in a general Pseudo-riemannian space.
3. Below I am trying to calculate explicitly the adjoint for $\square_{g}$ in $H^{1}({{\cal{U}}_{t_{1}}^{+}},\nu_{g}=\sqrt{-g}d^{4}x)$. Any comments will be appreciated.

The wave operator is self-adjoint in $L^{2}({{\cal{U}}_{t_{1}}^{+}},\nu_{g})$ which means that for all $\psi,\omega \in C_{c}^{\infty}({{\cal{U}}_{t_{1}}^{+}})$ it is true that: $$(\psi,\square\omega)_{L^2}=(\square\psi,\omega)_{L^2}$$.

which allow us to write:

$$\int_{{{\cal{U}}_{t_{1}}^{+}}} \square_{g}\psi\omega \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} \psi\square_{g}\omega \nu_{g}$$

Now taking into account the contracted Ricci identity:

$$\square(\psi_{,i})=(\square\psi)_{,i}+R^{j}_{i}\psi_{,j}$$

we have the following equality:

$$ \int_{{{\cal{U}}_{t_{1}}^{+}}} (\square_{g}\psi)_{,i}\omega_{,i} \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} (\square_{g}(\psi_{,i})-R^{j}_{i}\psi_{,j})\omega_{,i} \nu_{g} $$

Now the first term using the self adjointness of $\square_{g}$ can be rewritten as: $$ \int_{{{\cal{U}}_{t_{1}}^{+}}}\square_{g}(\psi_{,i})\omega_{,i} \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} \psi_{,_{i}}\square_{g}(\omega_{,i}) \nu_{g} $$ which again using the contracted Ricci identities gives: $$ \int_{{{\cal{U}}_{t_{1}}^{+}}}\psi_{,_{i}}\square_{g}(\omega_{,i}) \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} \psi_{,_{i}}((\square_{g}\omega)_{,i}+R^{j}_{i}\omega_{,j}) \nu_{g} $$

Now putting together the above equalities we have that:

$$ \int_{{{\cal{U}}_{t_{1}}^{+}}}\square\psi\omega \nu_{g}+ \int_{{{\cal{U}}_{t_{1}}^{+}}}(\square\psi)_{i}\omega_{i} \nu_{g}=\int_{{{\cal{U}}_{t_{1}}^{+}}}\psi\square\omega \nu_{g}+ \int_{{{\cal{U}}_{t_{1}}^{+}}}\psi_{i}(\square\omega)_{i} \nu_{g}+\int_{{{\cal{U}}_{t_{1}}^{+}}}(R^{i}_{j}-R^{j}_{i})\psi_{,j}\omega_{,i} \nu_{g} $$

It seems that in the case $R^{i}_{j}=0$ then the $\square_{g}$ is self adjoint. Is that correct?

So to sum up.

My main question is:

How can I calculate the adjoint of the wave operator $\square_{g}$ in Sobolev Spaces?

I have a three part question, which I could only received an answer for the first part here.

1. The Laplace-Beltrami operator is an operator which is the typical example of a self-adjoint operator in $L^{2}$. I am wondering if this is also true for other Hilbert spaces $H^{k}$.
2. The second part of the question and in which I am more interested is the following: How much the results are true for metrics with are not positive definitive. For example, the D'Alambertian(wave operator) in a general Pseudo-riemannian space.
3. Below I am trying to calculate explicitly the adjoint for $\square_{g}$ in $H^{1}({{\cal{U}}_{t_{1}}^{+}},\nu_{g}=\sqrt{-g}d^{4}x)$. Any comments will be appreciated.

The wave operator is self-adjoint in $L^{2}({{\cal{U}}_{t_{1}}^{+}},\nu_{g})$ which means that for all $\psi,\omega \in C_{c}^{\infty}({{\cal{U}}_{t_{1}}^{+}})$ it is true that: $$(\psi,\square\omega)_{L^2}=(\square\psi,\omega)_{L^2}$$.

which allow us to write:

$$\int_{{{\cal{U}}_{t_{1}}^{+}}} \square_{g}\psi\omega \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} \psi\square_{g}\omega \nu_{g}$$

Now taking into account the contracted Ricci identity:

$$\square(\psi_{,i})=(\square\psi)_{,i}+R^{j}_{i}\psi_{,j}$$

we have the following equality:

$$ \int_{{{\cal{U}}_{t_{1}}^{+}}} (\square_{g}\psi)_{,i}\omega_{,i} \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} (\square_{g}(\psi_{,i})-R^{j}_{i}\psi_{,j})\omega_{,i} \nu_{g} $$

Now the first term using the self adjointness of $\square_{g}$ can be rewritten as: $$ \int_{{{\cal{U}}_{t_{1}}^{+}}}\square_{g}(\psi_{,i})\omega_{,i} \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} \psi_{,_{i}}\square_{g}(\omega_{,i}) \nu_{g} $$ which again using the contracted Ricci identities gives: $$ \int_{{{\cal{U}}_{t_{1}}^{+}}}\psi_{,_{i}}\square_{g}(\omega_{,i}) \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} \psi_{,_{i}}((\square_{g}\omega)_{,i}+R^{j}_{i}\omega_{,j}) \nu_{g} $$

Now putting together the above equalities we have that:

$$ \int_{{{\cal{U}}_{t_{1}}^{+}}}\square\psi\omega \nu_{g}+ \int_{{{\cal{U}}_{t_{1}}^{+}}}(\square\psi)_{i}\omega_{i} \nu_{g}=\int_{{{\cal{U}}_{t_{1}}^{+}}}\psi\square\omega \nu_{g}+ \int_{{{\cal{U}}_{t_{1}}^{+}}}\psi_{i}(\square\omega)_{i} \nu_{g}+\int_{{{\cal{U}}_{t_{1}}^{+}}}(R^{i}_{j}-R^{j}_{i})\psi_{,j}\omega_{,i} \nu_{g} $$

It seems that in the case $R^{i}_{j}=0$ then the $\square_{g}$ is self adjoint. Is that correct?

So to sum up.

My main question is:

How can I calculate the adjoint of the wave operator $\square_{g}$ in Sobolev Spaces?

I have a three part question, which I could only received an answer for the first part here.

1. The Laplace-Beltrami operator is an operator which is the typical example of a self-adjoint operator in $L^{2}$. I am wondering if this is also true for other Hilbert spaces $W^{k,2}$.
2. The second part of the question and in which I am more interested is the following: How much the results are true for metrics with are not positive definitive. For example, the D'Alambertian(wave operator) in a general Pseudo-riemannian space.
3. Below I am trying to calculate explicitly the adjoint for $\square_{g}$ in $H^{1}({{\cal{U}}_{t_{1}}^{+}},\nu_{g}=\sqrt{-g}d^{4}x)$. Any comments will be appreciated.

The wave operator is self-adjoint in $L^{2}({{\cal{U}}_{t_{1}}^{+}},\nu_{g})$ which means that for all $\psi,\omega \in C_{c}^{\infty}({{\cal{U}}_{t_{1}}^{+}})$ it is true that: $$(\psi,\square\omega)_{L^2}=(\square\psi,\omega)_{L^2}$$.

which allow us to write:

$$\int_{{{\cal{U}}_{t_{1}}^{+}}} \square_{g}\psi\omega \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} \psi\square_{g}\omega \nu_{g}$$

Now taking into account the contracted Ricci identity:

$$\square(\psi_{,i})=(\square\psi)_{,i}+R^{j}_{i}\psi_{,j}$$

we have the following equality:

$$ \int_{{{\cal{U}}_{t_{1}}^{+}}} (\square_{g}\psi)_{,i}\omega_{,i} \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} (\square_{g}(\psi_{,i})-R^{j}_{i}\psi_{,j})\omega_{,i} \nu_{g} $$

Now the first term using the self adjointness of $\square_{g}$ can be rewritten as: $$ \int_{{{\cal{U}}_{t_{1}}^{+}}}\square_{g}(\psi_{,i})\omega_{,i} \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} \psi_{,_{i}}\square_{g}(\omega_{,i}) \nu_{g} $$ which again using the contracted Ricci identities gives: $$ \int_{{{\cal{U}}_{t_{1}}^{+}}}\psi_{,_{i}}\square_{g}(\omega_{,i}) \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} \psi_{,_{i}}((\square_{g}\omega)_{,i}+R^{j}_{i}\omega_{,j}) \nu_{g} $$

Now putting together the above equalities we have that:

$$ \int_{{{\cal{U}}_{t_{1}}^{+}}}\square\psi\omega \nu_{g}+ \int_{{{\cal{U}}_{t_{1}}^{+}}}(\square\psi)_{i}\omega_{i} \nu_{g}=\int_{{{\cal{U}}_{t_{1}}^{+}}}\psi\square\omega \nu_{g}+ \int_{{{\cal{U}}_{t_{1}}^{+}}}\psi_{i}(\square\omega)_{i} \nu_{g}+\int_{{{\cal{U}}_{t_{1}}^{+}}}(R^{i}_{j}-R^{j}_{i})\psi_{,j}\omega_{,i} \nu_{g} $$

It seems that in the case $R^{i}_{j}=0$ then the $\square_{g}$ is self adjoint. Is that correct?

So to sum up.

My main question is:

How can I calculate the adjoint of the wave operator $\square_{g}$ in Sobolev Spaces?

I have a three part question, which I could only received an answer for the first part here.

1. The Laplace-Beltrami operator is an operator which is the typical example of a self-adjoint operator in $L^{2}$. I am wondering if this is also true for other Hilbert spaces $H^{k}$.
2. The second part of the question and in which I am more interested is the following: How much the results are true for metrics with are not positive definitive. For example, the D'Alambertian(wave operator) in a general Pseudo-riemannian space.
3. Below I am trying to calculate explicitly the adjoint for $\square_{g}$ in $H^{1}({{\cal{U}}_{t_{1}}^{+}},\nu_{g}=\sqrt{-g}d^{4}x)$. Any comments will be appreciated.

The wave operator is self-adjoint in $L^{2}({{\cal{U}}_{t_{1}}^{+}},\nu_{g})$ which means that for all $\psi,\omega \in C_{c}^{\infty}({{\cal{U}}_{t_{1}}^{+}})$ it is true that: $$(\psi,\square\omega)_{L^2}=(\square\psi,\omega)_{L^2}$$.

which allow us to write:

$$\int_{{{\cal{U}}_{t_{1}}^{+}}} \square_{g}\psi\omega \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} \psi\square_{g}\omega \nu_{g}$$

Now taking into account the contracted Ricci identity:

$$\square(\psi_{,i})=(\square\psi)_{,i}+R^{j}_{i}\psi_{,j}$$

we have the following equality:

$$ \int_{{{\cal{U}}_{t_{1}}^{+}}} (\square_{g}\psi)_{,i}\omega_{,i} \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} (\square_{g}(\psi_{,i})-R^{j}_{i}\psi_{,j})\omega_{,i} \nu_{g} $$

Now the first term using the self adjointness of $\square_{g}$ can be rewritten as: $$ \int_{{{\cal{U}}_{t_{1}}^{+}}}\square_{g}(\psi_{,i})\omega_{,i} \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} \psi_{,_{i}}\square_{g}(\omega_{,i}) \nu_{g} $$ which again using the contracted Ricci identities gives: $$ \int_{{{\cal{U}}_{t_{1}}^{+}}}\psi_{,_{i}}\square_{g}(\omega_{,i}) \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} \psi_{,_{i}}((\square_{g}\omega)_{,i}+R^{j}_{i}\omega_{,j}) \nu_{g} $$

Now putting together the above equalities we have that:

$$ \int_{{{\cal{U}}_{t_{1}}^{+}}}\square\psi\omega \nu_{g}+ \int_{{{\cal{U}}_{t_{1}}^{+}}}(\square\psi)_{i}\omega_{i} \nu_{g}=\int_{{{\cal{U}}_{t_{1}}^{+}}}\psi\square\omega \nu_{g}+ \int_{{{\cal{U}}_{t_{1}}^{+}}}\psi_{i}(\square\omega)_{i} \nu_{g}+\int_{{{\cal{U}}_{t_{1}}^{+}}}(R^{i}_{j}-R^{j}_{i})\psi_{,j}\omega_{,i} \nu_{g} $$

It seems that in the case $R^{i}_{j}=0$ then the $\square_{g}$ is self adjoint. Is that correct?

So to sum up.

My main question is:

How can I calculate the adjoint of the wave operator $\square_{g}$ in Sobolev Spaces?

I have a three part question, which I could only received an answer for the first part here.

1. The Laplace-Beltrami operator is an operator which is the typical example of a self-adjoint operator in $L^{2}$. I am wondering if this is also true for other Hilbert spaces $W^{k,2}$.
2. The second part of the question and in which I am more interested is the following: How much the results are true for metrics with are not positive definitive. For example, the D'Alambertian(wave operator) in a general Pseudo-riemannian space.
3. Below I am trying to calculate explicitly the adjoint for $\square_{g}$ in $H^{1}({{\cal{U}}_{t_{1}}^{+}},\nu_{g})$. Any comments will be appreciated.

The wave operator is self-adjoint in $L^{2}({{\cal{U}}_{t_{1}}^{+}},\nu_{g})$ which means that for all $\psi,\omega \in C_{c}^{\infty}({{\cal{U}}_{t_{1}}^{+}})$ it is true that: $$(\psi,\square\omega)_{L^2}=(\square\psi,\omega)_{L^2}$$.

which allow us to write:

$$\int_{{{\cal{U}}_{t_{1}}^{+}}} \square_{g}\psi\omega \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} \psi\square_{g}\omega \nu_{g}$$

Now taking into account the contracted Ricci identity:

$$\square(\psi_{,i})=(\square\psi)_{,i}+R^{j}_{i}\psi_{,j}$$

we have the following equality:

$$ \int_{{{\cal{U}}_{t_{1}}^{+}}} (\square_{g}\psi)_{,i}\omega_{,i} \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} (\square_{g}(\psi_{,i})-R^{j}_{i}\psi_{,j})\omega_{,i} \nu_{g} $$

Now the first term using the self adjointness of $\square_{g}$ can be rewritten as: $$ \int_{{{\cal{U}}_{t_{1}}^{+}}}\square_{g}(\psi_{,i})\omega_{,i} \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} \psi_{,_{i}}\square_{g}(\omega_{,i}) \nu_{g} $$ which again using the contracted Ricci identities gives: $$ \int_{{{\cal{U}}_{t_{1}}^{+}}}\psi_{,_{i}}\square_{g}(\omega_{,i}) \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} \psi_{,_{i}}((\square_{g}\omega)_{,i}+R^{j}_{i}\omega_{,j}) \nu_{g} $$

Now putting together the above equalities we have that:

$$ \int_{{{\cal{U}}_{t_{1}}^{+}}}\square\psi\omega \nu_{g}+ \int_{{{\cal{U}}_{t_{1}}^{+}}}(\square\psi)_{i}\omega_{i} \nu_{g}=\int_{{{\cal{U}}_{t_{1}}^{+}}}\psi\square\omega \nu_{g}+ \int_{{{\cal{U}}_{t_{1}}^{+}}}\psi_{i}(\square\omega)_{i} \nu_{g}+\int_{{{\cal{U}}_{t_{1}}^{+}}}(R^{i}_{j}-R^{j}_{i})\psi_{,j}\omega_{,i} \nu_{g} $$

It seems that in the case $R^{i}_{j}=0$ then the $\square_{g}$ is self adjoint. Is that correct?

So to sum up.

My main question is:

How can I calculate the adjoint of the wave operator $\square_{g}$ in Sobolev Spaces?

I have a three part question, which I could only received an answer for the first part here.

1. The Laplace-Beltrami operator is an operator which is the typical example of a self-adjoint operator in $L^{2}$. I am wondering if this is also true for other Hilbert spaces $W^{k,2}$.
2. The second part of the question and in which I am more interested is the following: How much the results are true for metrics with are not positive definitive. For example, the D'Alambertian(wave operator) in a general Pseudo-riemannian space.
3. Below I am trying to calculate explicitly the adjoint for $\square_{g}$ in $H^{1}({{\cal{U}}_{t_{1}}^{+}},\nu_{g}=\sqrt{-g}d^{4}x)$. Any comments will be appreciated.

The wave operator is self-adjoint in $L^{2}({{\cal{U}}_{t_{1}}^{+}},\nu_{g})$ which means that for all $\psi,\omega \in C_{c}^{\infty}({{\cal{U}}_{t_{1}}^{+}})$ it is true that: $$(\psi,\square\omega)_{L^2}=(\square\psi,\omega)_{L^2}$$.

which allow us to write:

$$\int_{{{\cal{U}}_{t_{1}}^{+}}} \square_{g}\psi\omega \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} \psi\square_{g}\omega \nu_{g}$$

Now taking into account the contracted Ricci identity:

$$\square(\psi_{,i})=(\square\psi)_{,i}+R^{j}_{i}\psi_{,j}$$

we have the following equality:

$$ \int_{{{\cal{U}}_{t_{1}}^{+}}} (\square_{g}\psi)_{,i}\omega_{,i} \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} (\square_{g}(\psi_{,i})-R^{j}_{i}\psi_{,j})\omega_{,i} \nu_{g} $$

Now the first term using the self adjointness of $\square_{g}$ can be rewritten as: $$ \int_{{{\cal{U}}_{t_{1}}^{+}}}\square_{g}(\psi_{,i})\omega_{,i} \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} \psi_{,_{i}}\square_{g}(\omega_{,i}) \nu_{g} $$ which again using the contracted Ricci identities gives: $$ \int_{{{\cal{U}}_{t_{1}}^{+}}}\psi_{,_{i}}\square_{g}(\omega_{,i}) \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} \psi_{,_{i}}((\square_{g}\omega)_{,i}+R^{j}_{i}\omega_{,j}) \nu_{g} $$

Now putting together the above equalities we have that:

$$ \int_{{{\cal{U}}_{t_{1}}^{+}}}\square\psi\omega \nu_{g}+ \int_{{{\cal{U}}_{t_{1}}^{+}}}(\square\psi)_{i}\omega_{i} \nu_{g}=\int_{{{\cal{U}}_{t_{1}}^{+}}}\psi\square\omega \nu_{g}+ \int_{{{\cal{U}}_{t_{1}}^{+}}}\psi_{i}(\square\omega)_{i} \nu_{g}+\int_{{{\cal{U}}_{t_{1}}^{+}}}(R^{i}_{j}-R^{j}_{i})\psi_{,j}\omega_{,i} \nu_{g} $$

It seems that in the case $R^{i}_{j}=0$ then the $\square_{g}$ is self adjoint. Is that correct?

So to sum up.

My main question is:

How can I calculate the adjoint of the wave operator $\square_{g}$ in Sobolev Spaces?